[洛谷P4092][HEOI2016/TJOI2016]树
2024-09-03 23:27:38
题目大意:给你一棵树,有两个操作:
- $C\;x:$给第$x$个节点打上标记
- $Q\;x:$询问第$x$个节点的祖先中最近的打过标记的点(自己也是自己的祖先)
题解:树剖,可以维护区间或,然后若一段区间为$0$则跳过,否则在线段树上二分
卡点:二分部分多大了一个$=$,然后$MLE$
C++ Code:
#include <cstdio>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cstdlib>
namespace __IO {
namespace R {
int x, ch;
inline int read() {
ch = getchar();
while (isspace(ch)) ch = getchar();
for (x = ch & 15, ch = getchar(); isdigit(ch); ch = getchar()) x= x * 10 + (ch & 15);
return x;
}
inline char readc() {
ch = getchar();
while (!isalpha(ch)) ch = getchar();
return static_cast<char> (ch);
}
}
}
using __IO::R::read;
using __IO::R::readc; #define maxn 100010
int head[maxn], cnt;
struct Edge {
int to, nxt;
} e[maxn << 1];
inline void add(int a, int b) {
e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
} int n, m; int dfn[maxn], idx, fa[maxn], sz[maxn];
int son[maxn], top[maxn], dep[maxn], ret[maxn];
void dfs1(int u) {
sz[u] = 1;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa[u]) {
dep[v] = dep[u] + 1;
fa[v] = u;
dfs1(v);
if (!son[u] || sz[v] > sz[son[u]]) son[u] = v;
sz[u] += sz[v];
}
}
}
void dfs2(int u) {
dfn[u] = ++idx, ret[idx] = u;
int v = son[u];
if (v) top[v] = top[u], dfs2(v);
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != son[u] && v != fa[u]) {
top[v] = v;
dfs2(v);
}
}
} namespace SgT {
bool V[maxn << 2];
int pos, L, R; void modify(int rt, int l, int r) {
V[rt] = true;
if (l == r) return ;
int mid = l + r >> 1;
if (pos <= mid) modify(rt << 1, l, mid);
else modify(rt << 1 | 1, mid + 1, r);
}
void modify(int __pos) {
pos = __pos;
modify(1, 1, n);
} bool res;
void query(int rt, int l, int r) {
if (L <= l && R >= r) return static_cast<void> (res |= V[rt]);
int mid = l + r >> 1;
if (L <= mid) query(rt << 1, l, mid);
if (res) return ;
if (R > mid) query(rt << 1 | 1, mid + 1, r);
}
bool query(int __L, int __R) {
res = false;
L = __L, R = __R;
query(1, 1, n);
return res;
} int ans;
void ask(int rt, int l, int r) {
if (!V[rt]) return ;
if (l == r) {
if (!ans) ans = l;
return ;
}
int mid = l + r >> 1;
if (R > mid) ask(rt << 1 | 1, mid + 1, r);
if (ans) return ;
if (L <= mid) ask(rt << 1, l, mid);
}
int ask(int __L, int __R) {
L = __L, R = __R;
ans = 0;
ask(1, 1, n);
return ret[ans];
}
} int query(int x) {
while (top[x] != 1) {
if (SgT::query(dfn[top[x]], dfn[x])) return SgT::ask(dfn[top[x]], dfn[x]);
x = fa[top[x]];
}
return SgT::ask(1, dfn[x]);
} int main() {
n = read(), m = read();
for (int i = 1, a, b; i < n; i++) {
a = read(), b = read();
add(a, b);
}
dfs1(1);
top[1] = 1;
dfs2(1); SgT::modify(1);
while (m --> 0) {
char op = readc();
int x = read();
if (op == 'C') {
SgT::modify(dfn[x]);
} else {
printf("%d\n", query(x));
}
}
return 0;
}
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